why cant we pass &array to function where &array[0] is possible

Question

void fun(int* array){}  


int main(){

int array[]={1,2,3};
fun(&array);----->(1)//error in this line
return 0;
}

error: cannot convert â

Answer 1

While the address of any aggregate (this means arrays and standard-layout classes) is guaranteed to be the same as the address of the first element, the types are different, and there is no implicit conversion.

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As an example why the different type is important and useful:

for( int *p = array; p < (&array)[1]; ++p )

iterates over all elements of array, while

for( int *p = array; p < (&array[0])+1; ++p )

only executes once.